HM LI, LM. HM BN LI, BN LM

V NGUYN L BINThc hin: Phan nh Trung

I. nh ngha:

1. Hm li, lm:Hm s f c gi l li trn on [;] R nu vi mi x, y [;] v vi mi a, b 0 thaa+ b = 1 th:

f(ax+ by) af(x) + bf(y)

Ngc li, f c gi l lm trn on [;] R nu vi mi x, y [;] v vi mi a, b 0 thaa+ b = 1 th:

f(ax+ by) af(x) + bf(y)

Ch : Ta tha nhn kt qu sau:Nu hm f kh vi hai ln trn [;] th f li trn [;] khi f 0, x [;]. Ngc li, f lm trn[;] khi f 0, x [;]

2. Hm bn li, bn lm:Hm s f c gi l bn li, bn lm trn [;] R nu tn ti hng s [;] duy nht sao chof li trn [; ] v lm trn [;] hoc ngc li.

II. Tnh cht:

1. Nu f l hm li trn [;] th ta c: min f = min{f(); f()}

2. Nu f lm trn [;] th ta c: max f = max{f(); f()}

3. Nu f l mt hm lm trn on [;] th khi vi mi x, y, z R tha mn ng thi x z y v x+ y z th

f(x) + f(y) f(z) f(x+ y z)

Chng minh: t t = x z, d thy 0 t x y t = k(x y), k [0; 1]. Ta c:

f(x) + f(y) f(z) f(x+ y z) = f(x) + f(y) f(x t) f(y + t)= f(x) + f(y) f (x k(x y)) f (y + k(x y))= f(x) + f(y) f ((1 k)x+ ky) f (kx+ (1 k)y) f(x) + f(y) (1 k)f(x) kf(y) kf(x) (1 k)f(y) = 0 (Q.E.D)

4. Nu f li trn trn on [;] th khi vi mi x, y, z R tha mn ng thi x z y v x+ y z th

f(x) + f(y) f(z) f(x+ y z)

5. Nu f lm trn [;] th vi dy s xn [;] than

i=1 xi + (n 1) th khi :

f(x1) + f(x2) + ...+ f(xn) (n 1)f(a) f(x1 + x2 + ...+ xn (n 1))

Chng minh: Ta chng minh bng qui np theo nVi n = 1, hin nhin ng.Gi s tnh cht 5 ng ti n. Ta xt khi n+ 1.Khng gim tnh tng qut, gi s xn+1 = max{x1;x2; ...;xn+1}. S dng gi thit qui np ta ch cnchng minh:

f(x1 + x2 + ...+ xn (n 1)) + f(xn+1) f() + f(x1 + x2 + ...+ xn+1 n)

1

Do xn+1 = max{x1, x2, ..., xn+1} v xi [;] nn:

f() + f(x1 + x2 + ...+ xn+1 n) f(x1 + x2 + ...+ xn+1 n+ xn+1) + f(xn+1)= f(x1 + x2 + ...+ xn (n 1)) + f(xn+1)

Nh vy, khng nh ng vi n+ 1. Ta c pcm.

6. Nu f li trn [;] th vi dy s xn [;] than

i=1 xi + (n 1) th khi :

f(x1) + f(x2) + ...+ f(xn) (n 1)f(a) f(x1 + x2 + ...+ xn (n 1))

7. (BDT tip tuyn) Nu f lm trn [;] th vi mi x, x0 [;] ta c :

f(x) f (x0)(x x0) + f(x0)

Nu f li trn [;] th vi mi x, x0 [;] ta c :

f(x) f (x0)(x x0) + f(x0)

c bit, khi f(x) = ax+ b, a 6= 0 th th f l mt on thng nn:

min{f(); f()} f(x) max{f(); f()}

nh l 1: Cho (xn) l dy cc s thc tha:i) (xn) l dy khng gim.ii) xn [;]iii) x1 + x2 + ...+ xn = k = constiv) f l hm trn [;] tha f li trn [; ] v lm trn[;].

t F = f(x1) + f(x2) + ...+ f(xn). Khi :+ minF x1 = x2 = ... = xk1 = a;xk+1 = ... = xn [;]+ maxF x1 = x2 = ... = xk1 [;];xk+1 = ... = xn = b

nh l 2: Cho (xn) l dy cc s thc tha:i) (xn) l dy khng gim.ii) xn [;]iii) x1 + x2 + ...+ xn = k = constiv) f l hm trn [;] tha f lm trn [; ] v li trn[;].

t F = f(x1) + f(x2) + ...+ f(xn). Khi :+ maxF x1 = x2 = ... = xk1 = a;xk+1 = ... = xn [;]+ minF x1 = x2 = ... = xk1 [;];xk+1 = ... = xn = b

nh l 3: Cho (xn) l dy cc s thc tha:i) (xn) l dy khng gim.ii) xn Riii) x1 + x2 + ...+ xn = k = constiv) f l hm trn [;] tha f li trn (; ] v lm trn[; +).

t F = f(x1) + f(x2) + ...+ f(xn). Khi :+ minF x1 x2 = x3 = ...xn+ maxF x1 = x2 = ... = xn1 xn

nh l 4: Cho (xn) l dy cc s thc tha:i) (xn) l dy khng gim.

2

ii) xn Riii) x1 + x2 + ...+ xn = k = constiv) f l hm trn [;] tha f lm trn (; ] v li trn[; +).

t F = f(x1) + f(x2) + ...+ f(xn). Khi :+ maxF x1 x2 = x3 = ...xn+ minF x1 = x2 = ... = xn1 xn

III. Cc v d:

V d 1: (IMO Shorlist 1993, Viet Nam) Cho a, b, c, d l cc s thc khng m tha a+b+c+d = 1.CMR:

abc+ abd+ acd+ bcd 127

+176

27abcd

Li gii:

Gi s a = min{a, b, c, d} a 14. BDT cn chng minh vit li di dng:

(bc+ cd+ bd 176

27bcd

)a+ bcd 1

27(1)

Xem VT ca (1) nh mt hm s theo a. Ta xt cc trng hp: Trng hp 1: bc+cd+bd17627

bcd = 0.

Khi :

V T(1) = bcd (b+ c+ d)3

27 (a+ b+ c+ d)

3

27=

1

27

Trng hp 2: bc+ cd+ bd 17627

bcd < 0. Khi :

V T(1) bcd 1

27

Trng hp 3: bc+ cd+ bd 17627

bcd > 0. Khi :

V T(1) bc+ cd+ bd

4 17

27bcd

S dng BDT Schur, ta c:

bc+ cd+ bd

4 17

27bcd 13

324(bc+ cd+ bd) +

17

576 13(b+ c+ d)

2

324.3+

17

576=

1

27

Php chng minh hon tt.

Lu : Ty theo bi ton ra cn la i lng thay i ph hp sao cho thun li trong viccng ph bi ton. i vi bi ton ny, ta c th c nh c, d th do a+ b+ c+ d = 1 nn a+ b cngc nh, lc ny kho st V T(1) theo ab ta cng thu dc pcm.

V d 2: (Vasile Cirtoaje) Cho a, b, c, d l cc s thc khng m a, b, c, d tha a2 + b2 + c2 + d2 = 1.CMR:

(1 a)(1 b)(1 c)(1 d) abcd

Li gii:

Gi s a = min{a, b, c, d} a 12. Ta vit BDT cn chng minh di dng:

[(1 b)(1 c)(1 d) + bcd]a (1 b)(1 c)(1 d) 0 (1)

3

Xem VT ca (1) nh mt hm theo a. Ta c hai trng hp:Trng hp 1: (1 b)(1 c)(1 d) + bcd = 0, hin nhin ng.Trng hp 2: (1 b)(1 c)(1 d) + bcd > 0. Khi :

[(1 b)(1 c)(1 d) + bcd]a (1 b)(1 c)(1 d) bcd (1 b)(1 c)(1 d)2

t

p = b+ c+ d

q = bc+ cd+ db

r = bcd

. Khi , ta cn CM:

2r + p q 12

0

Ch l lc ny p2 2q = 34nn ta cn CM:

4r 14 (p 1)2

Theo AM-GM: p 3 3r T y ch vic kho st hm g(r) = 4r9 3

r2+6

r 5

4, r 1

8l c pcm.

V d 3: (Crux Mathemmaticorum) Cho a, b, c l cc s khng m. CMR:1 +

48a

b+ c+

1 +

48b

c+ a+

1 +

48c

a+ b 15

Li gii:

t x =2a

b+ c; y =

2b

c+ a; z =

2c

a+ b. Khi , ta c:

xy + yz + zx+ xyz = 4

V ta cn CM: 1 + 24x+

1 + 24y +

1 + 24z 15 (1)

Khng gim tnh tng qut, gi s: x = max{x, y, z} x 1. Xem VT ca (1) nh mt hm f theo x.D thy f ng bin nn f t gi tr nh nht khi x = 1. Lc ny, ta c y+z+2yz = 4 y = 4 z

1 + 2zv ta cn tm GTNN ca

1 + 24y +1 + 24z =

1 + 24.

4 z1 + 2z

+1 + 24z

Thc hin kho st hm g(z) =

1 + 24.4 z1 + 2z

+1 + 24z vi z [0; 1] ta d dng thy g(z) 10

T suy ra: 1 + 24x+

1 + 24y +

1 + 24z 15 (Q.E.D)

V d 4: Cho cc s a, b, c khng m. CMR:(4a

b+ c+ 1

)(4b

c+ a+ 1

)(4c

a+ b+ 1

) 25

Li gii:

t x =2a

b+ c, y =

2b

c+ a, z =

2c

a+ b. Khi :

xy + yz + zx+ xyz = 4

V ta cn CM:(1 + 2x)(1 + 2y)(1 + 2z) 25

4

Nu ta xem VT ca BDT ny nh mt hm theo z th r rng y l mt hm bc nht v ng bin.Theo . GTNN ca hm ny t c khi z = 0. Lc ny xy = 4 v ta cn ch ra:

(1 + 2x)(1 + 2y) 25 x+ y 8

iu ny l hin nhin theo AM-GM. Bi ton c chng minh.

V d 5: (USA MO 2000) Cho a, b, c l cc s thc khng m. CMR:

a+ b+ c

3 3abc max{(

ab)2; (

bc)2; (

ca)2}

Li gii:Khng mt tnh tng qut, gi s a b c. Khi , ta cn CM:

a+ b+ c 3 3abc 3(

ca)2 0 (1)

Xem VT ca (1) l mt hm theo b. Ta tnh c

f (b) =3ca

33b5> 0

Do f l hm lm. Nh vy GTLN ca f t c khi b = a hoc b = c. Ta xt khi a = b. Khi ta cn CM:

a+ 2c+ 33ca2 6

ca

iu ny l ng theo AM-GM.

Nu m rng BDT trn ta tm c chn di ca i lng TBC-TBN l1

2min1i

Tng qut ha bi ton v d 6, ta thu c bi ton ca tp ch Crux Mathemmaticorum:Cho x1, x2, ..., xn (n 2) l cc s khng m c tng bng 1. CMR:

1 x11 + x1

+

1 x21 + x2

+ ...+

1 xn1 + xn

n 2 + 23

V d 7: (VMEO 2005) Cho a, b, c l cc s thc dng cho trc v x, y, z l cc s thc dng thamn ax+ by + cz = xyz. Tm GTNN ca:

P = x+ y + z

Li gii:B : Vi cc s thc dng a, b, c cho trc. Khi tn ti duy nht d > 0 sao cho:

1

a+ d+

1

b+ d+

1

c+ d=

2

d

Chng minh b : Vit biu thc trn li thnh:

a

a+ d+

b

b+ d+

c

c+ d= 1

Xem y l mt hm f theo d. D thy f lin tc v f nghch bin. Mt khc limd0 f(d) = 3 vlimd+ f(d) = 0 nn tn ti duy nht s d > 0 tha mn ng thc trn. B c chng minh.

Quay li bi ton. t x =

bc

am, y =

ca

bn, z =

ab

cp. Khi ta c:

m+ n+ p = mnp

iu ny gi cho ta chn A,B,C l ba gc ca mt tam gic nhn sao cho: m = tanA,n = tanB, p =tanC. By gi, ta cn tm GTNN ca:

P =

bc

atanA+

ca

btanB +

ca

btanC

Xt hm f(x) = tanx, x (0;

2

). Ta c : f(x) = 2 tanx.(tan2 x + 1) > 0, x

(0,

2

)nn f lm

trn(0;

2

). Ta chn d l s thc dng sao cho:

1

a+ d+

1

b+ d+

1

c+ d=

2

d

S tn ti ca d hon ton c khng nh theo b . p dng BDT tip tuyn, ta c:

f(A) f (arctan

a(b+ d)(c+ d)

bc(a+ d)

)[A arctan

a(b+ d)(c+ d)

bc(a+ d)

]+ f

(arctan

a(b+ d)(c+ d)

bc(a+ d)

)

tanA a(b+ d)(c+ d) + bc(a+ d)bc(a+ d)

[A arctan

a(b+ d)(c+ d)

bc(a+ d)

]+

a(b+ d)(c+ d)

bc(a+ d)

bc

atanA 2abc+ (ab+ bc+ ca)d+ ad

2

abc(a+ d)

[A arctan

a(b+ d)(c+ d)

bc(a+ d)

]+

(b+ d)(c+ d)

a+ d

Ch rng v1

a+ d+

1

b+ d+

1

c+ d=

2

d

nn ta c ngayd3 = 2abc+ (ab+ bc+ ca)d

6

Suy ra: bc

atanA d

2

abc

[A arctan

a(b+ d)(c+ d)

bc(a+ d)

]+

(b+ d)(c+ d)

a+ d

Tng t, ta cng c:ca

btanB d

2

abc

[B arctan

b(a+ d)(c+ d)

ac(b+ d)

]+

(a+ d)(c+ d)

b+ d

ab

ctanC d

2

abc

[C arctan

c(a+ d)(b+ d)

ab(c+ d)

]+

(a+ d)(b+ d)

c+ d

Ch rng:a(b+ d)(c+ d)

bc(a+ d)+

b(c+ d)(a+ d)

ca(b+ d)+

c(a+ d)(b+ d)

ab(c+ d)=

(a+ d)(b+ d)(c+ d)

abc

nn:

A+B + C arctan

a(b+ d)(c+ d)

bc(a+ d) arctan

b(c+ d)(a+ d)

ca(b+ d)+ arctan

c(a+ d)(b+ d)

ab(c+ d)= 0

Do :

P

(a+ d)(b+ d)

c+ d+

(b+ d)(c+ d)

a+ d+

(c+ d)(a+ d)

b+ d

Vy:

minP =

(b+ d)(c+ d)

a+ d+

(c+ d)(a+ d)

b+ d+

(a+ d)(b+ d)

c+ d

Kt thc bi vit, ti xin ngh mt s bi tp t luyn:1. Cho a, b, c, d [0; 1]. CMR:

(1 a)(1 b)(1 c)(1 d) + a+ b+ c+ d 1

2. (THTT 2002) Cho a, b, c l cc s thc c tng bnh phng bng 2. Tm GTLN ca

P = 3(a+ b+ c) 22abc

3. Cho a, b, c, d 0 tha a+ b+ c+ d = 4. CMR:

16 + 2abcd 3(ab+ ac+ ad+ bc+ bd+ cd)

4. (Viet Nam TST 2001, Trn Nam Dng) Cho a, b, c l cc s dng tha 21ab+2bc+8ca 12. TmGTNN ca:

P =1

a+

2

b+

3

c

5. (IMO 1984) Cho x, y, z 0 tha x+ y + z = 1. CMR:

0 xy + yz + zx 2xyz 727

6. Cho x, y, z > 0 tha x+ y + z = 1. Tm GTLN ca

F =x

x2 + 1+

y

y2 + 1+

z

z2 + 1

7. (VMEO 2004) Cho tam gic nhn ABC. Tm GTNN ca:

P = tanA+ 2 tanB + 5 tanC

7

8. Cho x, y, z, t 0 tha x+ y + z + t = 4. CMR:

(1 + x)(1 + 3y)(1 + 3z)(1 + 3t) 125 + 131xyzt

Ti liu tham kho:[1]. Bt ng thc- V Quc B Cn.[2]. K thut nh gi trn bin - Phm Quc Thng.[3]. Nguyn l bin - Cao nh Huy.[4]. www.diendantoanhoc.net

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